We Width of the channel (m)There is a channel into the fjord and the model will describe the water speed in and out of the fjord and the water level into the fjord. Thevariables and parameters included are given in Table 1 and Figure 1 and Figure 2 describe the area of the fjord model horizontally.Horizontal picture of the fjordVertical cross sectionUcHeWc71A2XTable of variables and parameters included in the problem.t Time (s)WM2 Half-day tidal frequency (rad s – 1)nM2 Half-day tidal water amplitude (m)ni Water level in the open sea outside the fjord (m)Water laun2 Water level inside the fjord (m)Uc Flow rate in the channel (ms )Wc Width of theHc Depth of channel (m)Ac Area of a cross-section across the channel (Ac = Wc x Hc) (m²)Lc Length of the channel (m)A2 Surface of the fjord (m?)TM2 Half-day tide period (TM2 = 12.42 hours and wM2 = 2n/TM2)g Gravity (g = 9.81 ms-2)The current through the channel is driven by changes in the sea level outside, nı, which is due to the tide. The equation for ni (t) is71 =NM2sin(wM2t) .(1)The changes in the water level in the fjord, n2 (t), depend on the area of the cross-section of the channel, Ac, the flow velocity in thechannel, Uc, and the surface area of the fjord, A2. Note that Uc>0 means that water flows out of the fjord, and that Uc <0 means thatwater flows into the fjord. The differential equation describing this is:dtA2(2)The changes in flow velocity in the channel depend on the differences in water levels on the inside and outside of the channel and onthe length of the channel, Lc. The differential equation describing this is:dUc12%3DdtLc(3)With given initial conditions for n2(t) and Uc(t), equations (1), (2) and (3) can be used to find exact solutions for these two variables.QUESTION:From equations (1), (2) and (3), a second-order differential equation for Uc can be made in the formU" + aUc = f(t).determine a and f(t) in the equation above

Given: η1=ηM2sinωM2t ---------(1) dη2dt=-ACUCA2 ---------(2) dUCdt=gη2-η1LC ---------(3)…

There is a channel into the fjord and the model will describe the water speed in and out of the fjord and the water level into the fjord. The variables and parameters included are given in Table 1 and Figure 1 and Figure 2 describe the area of the fjord model horizontally. Y Horizontal picture of the fjord Vertical cross section Uc Z-0 He Wc A2 Lc Table of variables and parameters included in the problem. t Time (s) Wm2 Half-day tidal frequency (rad s - 1) nM2 Half-day tidal water amplitude (m) ni Water level in the open sea outside the fjord (m) n2 Water level inside the fjord (m) Uc Flow rate in the channel (ms-!) Wc Width of the channel (m) Hc Depth of channel (m) Ac Area of a cross-section across the channel (Ac = Wc x Hc) (m²) Lc Length of the channel (m) A2 Surface of the fjord (m?) Tm2 Half-day tide period (T2 = 12.42 hours and wm2 = 21/Tm2) E Gravity (g = 9.81 ms-2) The current through the channel is driven by changes in the sea level outside, ni, which is due to the tide. The equation for ni (t) is Ni = ]M2sin(wM2t). (1) The changes in the water level in the fjord, n2 (t), depend on the area of the cross-section of the channel, Ac, the flow velocity in the channel, Uc, and the surface area of the fjord, A2. Note that Uc> 0 means that water flows out of the fjord, and that Uc <0 means that water flows into the fjord. The differential equation describing this is: dt (2) The changes in flow velocity in the channel depend on the differences in water levels on the inside and outside of the channel and on the length of the channel, Lc. The differential equation describing this is: dUc g- Lc dt (3) With given initial conditions for na(t) and Uc(t), equations (1), (2) and (3) can be used to find exact solutions for these two variables. QUESTION: From equations (1), (2) and (3), a second-order differential equation for Uc can be made in the form U" + aUc = f(t). determine a and f(t) in the equation above

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter7: Analytic Trigonometry

Section7.6: The Inverse Trigonometric Functions

Problem 94E